What is the solution for 4y^{''}+12y^'+9y=0,y(0)=1,y^'(0)=-5 ?

The solution for 4y^{''}+12y^'+9y=0,y(0)=1,y^'(0)=-5 is y=e^{-(3t)/2}-(7e^{-(3t)/2}t)/(2)

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4y^{''}+12y^'+9y=0,y(0)=1,y^'(0)=-5

4 y^{^{''}} + 12 y^{^{'}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = - 5

y = e^{- \frac{3 t}{2}} - \frac{7 e ^{- \frac{3 t}{2}} t}{2}

Solution steps

4 y^{''} (t) + 12 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = e^{- \frac{3 t}{2}} - \frac{7 e ^{- \frac{3 t}{2}} t}{2}

y = e^{- \frac{3 t}{2}} - \frac{7 e ^{- \frac{3 t}{2}} t}{2}

x sin (x)

f (x) = 1 + 8 x

(1 + x^{4})^{\frac{2}{3}}

\frac{d}{d x} (\frac{( 5 x - 3 )}{( 4 x - \frac{3}{x} )})

\frac{\partial}{\partial x} (y (- 14 x - 4 y + 1))

What is the solution for 4y^{''}+12y^'+9y=0,y(0)=1,y^'(0)=-5 ?
The solution for 4y^{''}+12y^'+9y=0,y(0)=1,y^'(0)=-5 is y=e^{-(3t)/2}-(7e^{-(3t)/2}t)/(2)